NCERT Solution class 7 chapter 2 exercise 2.7

Exercise 2.7

Question 1

Find:

(i) 0.4 \div 2

Sol :

0.4 \div 2 =\dfrac{4}{10} \div 2


(ii) 0.35 \div 5

Sol :

0.35 \div 5 =\dfrac{35}{100} \div 5 =\dfrac{35}{100} \times \dfrac{1}{5} =\dfrac{7}{100} = 0.07


(iii) 2.48 \div 4

Sol :

2.48 \div 4 =\dfrac{248}{100} \div 4 =\dfrac{248}{100} \times \dfrac{1}{4} =\dfrac{62}{100} = 0.62


(iv) 65.4 \div 6

Sol :

65.4 \div 6 =\dfrac{654}{10} \div 6 \dfrac{654}{10} \times \dfrac{1}{6} =\dfrac{109}{10} = 10.9


(v) 651.2 \div 4

Sol :

651.2 \div 4 =\dfrac{6512}{10} \div 4 =\dfrac{6512}{10} \times \dfrac{1}{4} =\dfrac{1628}{10} = 162.8


(vi) 14.49 \div 7

Sol :

14.49 \div 7 =\dfrac{1449}{100} \div 7 =\dfrac{1449}{100} \times \dfrac{1}{7} =\dfrac{207}{100} = 2.07


(vii) 3.96 \div 4

Sol :

3.96 \div 4 =\dfrac{396}{100} \div 4  =\dfrac{396}{100} \times \dfrac{1}{4} =\dfrac{99}{100} = 0.99


(viii) 0.80 \div 5

Sol :

0.80 \div 5 =\dfrac{80}{100} \div 5 =\dfrac{80}{100} \times \dfrac{1}{5} =\dfrac{16}{100} =0.16


 

Question 2

Find:

(i) 4.8 \div  10

Sol : 0.48


(ii) 52.5 \div 10

Sol : 5.25


(iii) 0.7 \div  10

Sol : 0.07


(iv) 33.1 \div 10

Sol : 3.31


(v) 272.23 \div 10

Sol : 27.223


(vi) 0.56 \div 10

Sol : 0.056


(vii) 397\div 10

Sol : 0.397


Note : We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 10, the decimal will shift to the left by 1 place.


 

Question 3

Find:
(i) 2.7 \div 100

Sol : 0.027


(ii) 0.3 \div 100

Sol : 0.003


(iii) 0.78 \div 100

Sol : 0.0078


(iv) 432.6 \div 100

Sol : 4.326


(v) 23.6 \div 100

Sol : 0.236


(vi) 98.53 \div 100

Sol : 0.9853


Notes : We know that when a decimal number is divided by a multiple of 10 only (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeroes. Since here we are dividing by 100, the decimal will shift to the left by 2 places.


 

Question 4

Find:
(i) 7.9 \div 1000

Sol : 0.0079


(ii) 26.3 \div 1000

Sol : 0.0263


(iii) 38.53 \div 1000

Sol : 0.03853


(iv) 128.9 \div 1000

Sol : 0.1289


(v) 0.5 \div 1000

Sol : 0.0005


 

Question 5

Find:

(i) 7 \div 3.5

Sol :

7 \div 3.5 =7 \div \dfrac{35}{10} =35 \times \dfrac{10}{35}  = 2


(ii) 36 \div 0.2

Sol :

36 \div 0.2 =36 \div \dfrac{2}{10} =36 \times \dfrac{10}{2}  = 180


(iii) 3.25 \div 0.5

Sol :

3.25 \div 0.5 =\dfrac{325}{100} \div \dfrac{5}{10} =\dfrac{325}{100} \times \dfrac{10}{5} =\dfrac{65}{10} = 6.5


(iv) 30.94\div 0.7

Sol :

30.94\div 0.7 =\dfrac{3094}{100} \div \dfrac{7}{10} =\dfrac{3094}{100} \times \dfrac{7}{10} =\dfrac{442}{10} = 44.2


(v) 0.5 \div 0.25

Sol :

0.5 \div 0.25 =\dfrac{5}{10} \div \dfrac{25}{100} =\dfrac{5}{10} \times \dfrac{100}{25}  = 2


(vi) 7.75 \div 0.25

Sol :

7.75 \div 0.25 =\dfrac{775}{100} \div \dfrac{25}{100} =\dfrac{775}{100} \times \dfrac{100}{25}  = 31


(vii) 76.5 \div 0.15

Sol :

76.5 \div 0.15 =\dfrac{765}{10} \div \dfrac{15}{100} =\dfrac{765}{10} \times \dfrac{100}{15} = 510


(viii) 37.8 \div 1.4

Sol :

37.8 \div 1.4 =\dfrac{378}{10} \div \dfrac{14}{10} =\dfrac{378}{10} \times \dfrac{10}{14}  = 27


(ix) 2.73 \div 1.3

Sol :

2.73 \div 1.3 =\dfrac{273}{100} \div \dfrac{13}{10} =\dfrac{273}{100} \times \dfrac{10}{13} =\dfrac{21}{10} = 2.1


 

Question 6

A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?

Sol :

Distance covered in 2.4 litres of petrol = 43.2 km

\therefore Distance covered in 1 litre of petrol =43.2\div 2.4 =\dfrac{432}{10} \div \dfrac{24}{10} =\dfrac{432}{10} \div \dfrac{10}{24} = 18

Therefore, the vehicle will cover 18 km in 1 litre petrol.


 

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